A simple perdulum performs simple harmonic motion about x=0 with … - Vidyadip

MCQ

A simple perdulum performs simple harmonic motion about $x=0$ with an amplitude $a$ and time period $T$. The speed of the pendulum at $x=a/2$ will be

A
$\frac{{\pi A}}{T}$
B
$\;\frac{{3{\pi ^2}A}}{T}$
✓
$\;\frac{{\pi A\sqrt 3 }}{T}$
D
$\;\frac{{\pi A\sqrt 3 }}{{2T}}$

✓

Answer

Correct option: C.

$\;\frac{{\pi A\sqrt 3 }}{T}$

c
For simple harmonic motion,

$v=\omega \sqrt{a^{2}-x^{2}}$

When $x=\frac{a}{2}, v=\omega \sqrt{a^{2}-\frac{a^{2}}{4}}=\omega \sqrt{\frac{3}{4} a^{2}}$

As $\omega=\frac{2 \pi}{T}, \quad \therefore \quad v=\frac{2 \pi}{T} \cdot \frac{\sqrt{3}}{2} a \Rightarrow \quad v=\frac{\pi \sqrt{3} a}{T}$

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